How do you find the Maclaurin Series for # (sin(x/11))^2#?

Question

1306 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-28T12:53:47

#sin^2 (x/11)= -1/2sum_(n=1)^oo (-1)^n (2/11)^(2n) x^(2n)/((2n)!)#

Use the trigonometric identity:

#sin^2 alpha = (1-cos(2alpha))/2#

Expand #cos(2alpha)# using the MacLaurin series for #cosx#:

#sin^2 alpha = 1/2-1/2sum_(n=0)^oo (-1)^n (2alpha)^(2n)/((2n)!)#

Extract the term for #n=0# from the series:

#sin^2 alpha = cancel(1/2)-cancel(1/2)-1/2sum_(n=1)^oo (-1)^n (2alpha)^(2n)/((2n)!)#

and finally substitute #alpha=x/11#:

#sin^2 (x/11)= -1/2sum_(n=1)^oo (-1)^n (2/11)^(2n) x^(2n)/((2n)!)#